Implicit Differentiation and Related Rates: Tangents to Circles and Beyond

In this video, the host introduces implicit differentiation through circle geometry and related rates, building intuition for how tiny changes in two interdependent variables relate to a derivative. The discussion moves from a circle x^2 + y^2 = 25 to a ladder sliding along a wall, illustrating how the same derivative concept governs different problems. The talk also connects implicit differentiation to broader ideas in multivariable calculus and ends with a look at derivatives of inverse functions.

• Understand how to compute slopes on implicit curves such as circles where x is not a function of y.

Introduction to Implicit Differentiation

The video introduces implicit differentiation as a method for finding slopes where the relationship between x and y is given by an equation in two variables rather than a function. The key idea is to think in terms of tiny nudges to both variables and to ask how a function of two variables changes as those nudges occur. This approach generalizes beyond functions to any implicit curve.

Circle Example and Implicit Slopes

Consider the circle of radius five centered at the origin described by x squared plus y squared equals 25. For a point on the circle, the tangent line is perpendicular to the radius. Differentiating both sides yields 2x dx plus 2y dy equals 0, which simplifies to dy/dx equals -x/y. At the point (3,4) this slope is -3/4. Here dx and dy are infinitesimal changes rather than a literal time step, which can feel unusual at first but is a natural way to describe instantaneous rates along an implicit curve.

Ladder Related Rates as a Related Problem

The ladder problem models a 5 meter long ladder against a wall with the top at a fixed height that changes over time. The relation x squared plus y squared equals 25 holds for all times, where x is the distance from the wall to the base and y is the height of the top above the ground. If the top slides downward at a rate of 1 meter per second, dy/dt equals -1, and solving the related derivatives gives dx/dt equals 4/3. This illustrates how the same geometric constraint links rate of change in two variables via time derivatives.

From Time to Two Variable Functions

The speaker then reframes the left side of the circle equation as a function of time, S equals x squared plus y squared. Since S is constant, its derivative with respect to time is zero. Differentiating 2x times dx/dt plus 2y times dy/dt equals zero recovers the same relationship between dx and dy as in the circle example. This viewpoint connects time based derivatives to the implicit differentiation framework.

General Rules with Two Variables

A second example uses a general implicit relation sin of x times y squared equals x. Applying the product rule yields a relation for dy in terms of dx, namely left side equals sine of x times 2y dy plus y squared times cos of x times dx. Equating the two sides and solving for dy/dx demonstrates how to extend implicit differentiation to a wide class of curves.

Logarithms and Inverse Functions

The derivative of the natural logarithm is derived via implicit differentiation by rewriting y equals ln x as e to the y equals x. Differentiating both sides gives e to the y times dy equals dx, so dy/dx equals 1 over e to the y. On the curve e to the y equals x, hence dy/dx equals 1 over x, which recovers the standard derivative of ln x.

Takeaways and Outlook

The talk ties implicit differentiation to broader multivariable calculus ideas, emphasizing the interpretation of tiny nudges and the role of partial changes. It also signals how limits will formalize the derivative in future discussions, highlighting the connection between visual geometric reasoning and analytic foundations.